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THE
METHOD OF ARCHIMEDES
RECENTLY DISCOVERED BY HEIBERG
A SUPPLEMENT TO THE WORKS
OF ARCHIMEDES 1897
EDITED BY
Sir THOMAS L. HEATH,
K.C.B., ScD., F.R.S.
SOMETIME FELLOIV OF TRINITY COLLEGE, CAMBRIDGE
Cambridge :
at the University Press
1912
Price Tivo Shillings and Sixpence net
THE
METHOD OF ARCHIMEDES
CA.MBRIDGE UNIVERSITY PRESS
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All rights reserved
THE
METHOD OF ARCHIMEDES
RECENTLY DISCOVERED BY HEIBERG
A SUPPLEMENT TO THE WORKS
OF ARCHIMEDES 1897
EDITED BY
Sir THOMAS L. HEATH, ^
K.C.B., Sc.D., F.R.S.
SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE
Cambridge :
at the University Press
1912
©amiiriiig? :
PRINTED BY JOHN CLAY, M.A
AT THE UNIVERSITY PRESS
150369
INTRODUCTORY NOTE
From the point of view of the student of Greek mathematics
there has been, in recent years, no event comparable in interest
with the discovery by Heiberg in 1906 of a Greek MS. containing,
among other works of Archimedes, substantially the whole of a
treatise which was formerly thought to be irretrievably lost.
The full description of the MS. as given in the preface to Vol. i.
(1910) of the new edition of Heibei'g's text of Archimedes now in
course of publication is —
Codex rescriptus Mefcochii Constantinopolitani S. Sepulchri
monasterii Hierosolymitani 355, 4to.
Heiberg has told the story of his discovery of this MS. and
given a full description of it*. His attention having been called
to a notice in Vol. iv. (1899) of the Ίΐροσολνμιηκη βιβλωθηκη of
Papadopulos Kerameus relating to a palimpsest of mathematical
content, he at once inferred from a few specimen lines which were
quoted that the MS. must contain something by Archimedes. As
the result of inspection, at Constantinople, of the MS. itself, and
by means of a photograph taken of it, he was able to see what it
contained and to decipher much of the contents. This was in the
year 1906, and he inspected the MS. once more in 1908. With
the exception of the last leaves, 178 to 185, which are of paper
of the 16th century, the MS. is of parchment and contains writings
of Archimedes copied in a good hand of the 10th century, in two
columns. An attempt was made (fortunately with only partial
success) to wash out the old writing, and then the parchment was
used again, for the purpose of writing a Euchologion thereon, in the
12th — 13th or 13th — 14th centuries. The earlier writing appears
with more or less clearness on most of the 177 leaves; only 29
leaves are destitute of any trace of such writing ; from 9 more
it was hopelessly washed off; on a few more leaves only a few
words can be made out; and again some 14 leaves have old writing
* Hermes xlii. 1907, pp. 235 sq.
6 INTRODUCTORY NOTE
upon them in a diflferent hand and with no division into columns.
All the rest is tolerably legible with the aid of a magnifying glass.
Of the treatises of Archimedes which are found in other MSS., the
new MS. contains, in great part, the books On the Sphere and
Cylinder, almost the whole of the work On Spirals, and some parts
of the Measurement of a Circle and of the books On the Equilibrium
of Planes. But the important fact is that it contains (1) a con-
siderable proportion of the work On Floating Bodies which was
formerly supposed to be lost so far as the Greek text is concerned
and only to have survived in the translation by Wilhelm von
Morbeke, and (2), most precious of all, the greater part of the
book called, according to its own heading, "Εφοδος and elsewhere,
alternatively, Έφόδιον or Έφοδικόν, meaning Method. The portion
of this latter work contained in the MS, has already been published
by Heiberg (1) in Greek* and (2) in a German translation with
commentary by Zeuthenf. The treatise was formerly only known
by an allusion to it in Suidas, who says that Theodosius wrote a
commentary upon it ; but the Metrica of Heron, newly discovered
by R. Schone and published in 1903, quotes three propositions from
\t%, including the two main propositions enunciated by Archimedes
at the beginning as theorems novel in character which the method
furnished a means of investigating. Lastly the MS. contains two
short propositions, in addition to the preface, of a work called
Stomachion (as it might be " Neck-Spiel " or " Qual-Geist ") which
treated of a sort of Chinese puzzle known afterwards by the name
of " loculus Archimedius " ; it thus turns out that this puzzle, which
Heiberg was formerly disinclined to attribute to Archimedes §, is
really genuine.
The Method, so happily recovered, is of the greatest interest for
the following reason. Nothing is more characteristic of the classical
works of the great geometers of Greece, or more tantalising, than
the absence of any indication of the steps by which they worked
their way to the discovery of their great theorems. As they have
come down to us, these theorems are finished masterpieces which
leave no traces of any rough-hewn stage, no hint of the method
by which they were evolved. We cannot but suppose that the
* Hermes XLn. 1907, pp. 243—297.
t Bibliotheca Mathematica VII3 , 1906-7, pp. 321 — 363.
ί Heronis Alexandrini ojjera, Vol. iii. 1903, pp. 80, 17 ; 130, 15 ; 130, 25.
§ Vide The Works of Archimedes, p. xxii.
INTEODUCTORY NOTE 7
Greeks had some method or methods of analysis hardly less powerful
than those of modern analysis ; yet, in general, they seem to have
taken pains to clear away all traces of the machinery used and all
the litter, so to speak, resulting from tentative efforts, before they
permitted themselves to publish, in sequence carefully thought out,
and with definitive and rigorously scientific proofs, the results
obtained. A partial exception is now furnished by the Method; for
here we have a sort of lifting of the veil, a glimpse of the interior
of Archimedes' workshop as it were. He tells us how he discovered
certain theorems in quadrature and cubature, and he is at the same
time careful to insist on the difference between (1) the means which
may be sufl&cient to suggest the truth of theorems, although not
furnishing scientific proofs of them, and (2) the rigorous demonstra-
tions of them by irrefragable geometrical methods which must follow
before they can be finally accepted as established ; to use Archi-
medes' own terms, the former enable theorems to be investigated
{OewpeLv) but not to be proved (άποΒεικννναή. The mechanical
method, then, used in our treatise and shown to be so useful for
the discovery of theorems is distinctly said to be incapable of
furnishing proofs of them ; and Archimedes promises to add, as
regards the two main theorems enunciated at the beginning, the
necessary supplement in the shape of the formal geometrical proof.
One of the two geometrical proofs is lost, but fragments of the other
are contained in the MS. which are sufficient to show that the
method was the orthodox method of exhaustion in the form in
which Archimedes applies it elsewhere, and to enable the proof to
be reconstructed.
The rest of this note will be best understood after the treatise
itself has been read ; but the essential features of the mechanical
method employed by Archimedes are these. Suppose X to be a
plane or solid figure, the area or content of which has to be found.
The method is to weigh infinitesimal elements of X (with or without
the addition of the corresponding elements of another figure C)
against the corresponding elements of a figure B, Β and C being
such figures that their areas or volumes, and the position of the
centre of gravity of B, are known beforehand. For this purpose
the figures are first placed in such a position that they have, as
common diameter or axis, one and the same straight line ; if then
the infinitesimal elements are sections of the figures made by parallel
planes perpendicular (in general) to the axis and cutting the figures,
8 INTRODUCTORY NOTE
the centres of gravity of all the elements lie at one point or other
on the common diameter or axis. This diameter or axis is produced
and is imagined to be the bar or lever of a balance. It is sufficient
to take the simple case where the elements of X alone are weighed
against the elements of another figure B. The elements which cor-
respond to one another are the sections of X and Β respectively by
any one plane perpendicular (in general) to the diameter or axis
and cutting both figures; the elements are spoken of as straight
lines in the case of plane figures and as plane areas in the case of
solid figures. Although Archimedes calls the elements straight lines
and plane areas respectively, they are of course, in the first case,
indefinitely narrow strips (areas) and, in the second case, indefinitely
thin plane laminae (solids) ; but the breadth or thickness {dx, as
we might call it) does not enter into the calculation because it is
regarded as the same in each of the two corresponding elements
which are separately weighed against each other, and therefore
divides out. The number of the elements in each figure is in-
finite, but Archimedes has no need to say this ; he merely says
that X and Β are imade up of all the elements in them respectively,
i.e. of the sti-aight lines in the case of areas and of the plane areas
in the case of solids.
The object of Archimedes is so to arrange the balancing of the
elements that the elements of X are all applied at one point of
the lever, while the elements of Β operate at difierent points,
namely where they actually are in the first instance. He con-
trives therefore to move the elements of X away from their first
position and to concentrate them at one point on the lever, while
the elements of Β are left where they are, and so operate at their
respective centres of gravity. Since the centre of gravity of Β as
a whole is known, as well as its area or volume, it may then be
supposed to act as one mass applied at its centre of gravity ; and
consequently, taking the whole bodies X and Β as ultimately placed
respectively, we know the distances of the two centres of gravity
from the fulcrum or point of suspension of the lever, and also the
area or volume of B. Hence the area or volume of X is found.
The method may be applied, conversely, to the problem of finding
the centre of gravity of X when its area or volume is known before-
hand ; in this case it is necessary that the elements of X, and
therefore X itself, should be weighed in the places where they are,
and that the figures the elements of which are moved to one sins;le
INTRODUCTORY NOTE 9
point of the lever, to be weighed there, should be other figures and
not X.
The method wUl be seen to be, not integration, as certain
geometrical proofs in the great treatises actually are, but a clever
device for avoiding the particular integration which would naturally
be used to find directly the area or volume required, and making
the solution depend, instead, upon another integration the result of
which is already known. Archimedes deals with moments about
the point of suspension of the lever, i.e. the products of the ele-
ments of area or volume into the distances between the point of
suspension of the lever and the centres of gravity of the elements
respectively ; and, as we said above, while these distances are
different for all the elements of £, he contrives, by moving the
elements of X, to make them the same for all the elements of X
in their final position. He assumes, as known, the fact that the
sum of the moments of each particle of the figure Β acting at
the point where it is placed is equal to the moment of the whole
figure applied as one mass at one point, its centre of gravity.
Suppose now that the element of X is u . dx, u being the length
or area of a section of X by one of a whole series of parallel planes
cutting the lever at right angles, χ being measured along the lever
(which is the common axis of the two figures) from the point of
suspension of the lever as origin. This element is then supposed
to be placed on the lever at a constant distance, say a, from the
origin and on the opposite side of it from B. If u' . dx is the cor-
responding element of Β cut off by the same plane and χ its distance
from the origin, Archimedes' argument establishes the equation
rk rk
a I udx = I xu'dx.
Jh Jh
Now the second integral is known because the area or volume of
the figure Β (say a triangle, a pyramid, a prism, a sphere, a cone,
or a cylinder) is known, and it can be supposed to be applied as
one mass at its centre of gravity, which is also known ; the integral
is equal to bU, where b is the distance of the centre of gravity from
the point of suspension of the lever, and U is the area or content
of B. Hence
the area or volume oi X = — .
a
In the case where the elements of X are weighed along with the
corresponding elements of another figure G against corresponding
10 INTRODUCTORY NOTE
elements of B, we have, if ν be the element of C, and V its area
or content,
rh rk rh
a I udx + a i vdx= j xu'dx
Jh Jh Jh
and (area or volume of X + V)a — hu.
In the particular problems dealt with in the treatise h is always
= 0, and k is often, but not always, equal to a.
Our admiration of the genius of the greatest mathematician of
antiquity must surely be increased, if that were possible, by a
perusal of the work before us. Mathematicians will doubtless
agree that it is astounding that Archimedes, writing (say) about
250 B.C., should have been able to solve such problems as those of
finding the volume and the centre of gravity of any segment of a
sphere, and the centre of gravity of a semicircle, by a method so
simple, a method too (be it observed) which would be quite rigorous
enough for us to-day, although it did not satisfy Archimedes himself.
Apart from the mathematical content of the book, it is in-
teresting, not only for Archimedes' explanations of the course which
his investigations took, but also for the allusion to Democritus as
the discoverer of the theorem that the volumes of a pyramid and
a cone are one-third of the volumes of a prism and a cylinder
respectively which have the same base and equal height. These
propositions had always been supposed to be due to Eudoxus, and
indeed Archimedes himself has a statement to this effect*. It
now appears that, though Eudoxus was the first to prove them
scientifically, Democritus was the first to assert their truth. I have
elsewhere t made a suggestion as to the probable course of Democritus'
argument, which, on Archimedes' view, did not amount to a proof
of the propositions ; but it may be well to re-state it here. Plutarch,
in a well-known passage |, speaks of Democritus as having raised the
following question in natural philosophy (φυσικώς) : "if a cone were
cut by a plane parallel to the base [by which is clearly meant a
plane indefinitely near to the base], what must we think of the
surfaces of the sections ? Are they equal or unequal ? For, if they
are unequal, they will make the cone irregular, as having many
indentations, like steps, and unevennesses ; but, if they are equal,
the sections will be equal, and the cone will appear to have the
property of the cylinder and to be made up of equal, not unequal,
* On the Sphere and Cylinder, Preface to Book i.
t The Thirteen Books of Euclid's Elements, Vol. iii. p. 368.
X Plutarch, De Comm. Not. adv. Stoicos xxxix. 3.
INTRODUCTORY NOTE 11
circles, which is very absurd." The phrase "made up of equal...
circles" (e^ ίσων σνγκζίμζνοζ . . . κνκλων) shows that Democritus already
had the idea of a solid being the sum of an infinite number of
parallel planes, or indefinitely thin laminae, indefinitely near to-
gether : a most important anticipation of the same thought which
led to such fruitful results in Archimedes. If then we may make
a conjecture as to Democritus' argument with regard to a pyramid,
it seems probable that he would notice that, if two pyramids of the
same height and with equal triangular bases ai'e respectively cut by
planes parallel to the base and dividing the heights in the same
ratio, the corresponding sections of the two pyramids are equal,
whence he would infer that the pyi-amids are equal because they
are the sums of the same infinite numbers of equal plane sections
or indefinitely thin laminae. (This would be a particular anti-
cipation of Cavalieri's proposition that the areal or solid contents
of two figures are equal if two sections of them taken at the same
height, whatever the height may be, always give equal sti^aight lines
or equal surfaces respectively.) And Democritus would of course
see that the three pyramids into wliich a prism on the same base
and of equal height with the original pyramid is divided (as in
Eucl. XII. 7) satisfy, in pairs, this test of equality, so that the
pyramid would be one tliird part of the prism. The extension to
a pyramid with a polygonal base would be easy. And Democritus
may have stated the proposition for the cone (of course without an
absolute proof) as a natural inference from the result of increasing
indefinitely the number of sides in a regular polygon forming the
base of a pyramid.
In accordance with the plan adopted in The Works of Archimedes,
I have marked by inverted commas the passages which, on account
of their importance, historically or otherwise, I have translated
literally from the Greek ; the rest of the tract is reproduced in
modern notation and phraseology. Words and sentences in square
brackets represent for the most part Heiberg's conjectural restoration
(in his German translation) of what may be supposed to have been
written in the places where the MS. is illegible; in a few cases
where the gap is considerable a note in brackets indicates what the
missing passage presumably contained and, so far as necessary, how
the deficiency may be made good.
T. L. H.
7 June 1912.
THE METHOD OF ARCHIMEDES TREATING
OF MECHANICAL PROBLEMS—
TO ERATOSTHENES
"Archimedes to Eratosthenes greeting.
I sent you on a former occasion some of the theorems
discovered by me, merely writing out the enunciations and
inviting you to discover the proofs, which at the moment
I did not give. The enunciations of the theorems which I
sent were as follows.
1. If in a right prism with a parallelogrammic base a
cylinder be inscribed which has its bases in the opposite
parallelograms*, and its sides [i.e. four generators] on the
remaining planes (faces) of the prism, and if through the
centre of the circle which is the base of the cylinder and
(through) one side of the square in the plane opposite to
it a plane be drawn, the plane so drawn will cut off from
the cylinder a segment which is bounded by two planes
and the surface of the cylinder, one of the two planes being
the plane which has been drawn and the other the plane
in which the base of the cylinder is, and the surface being
that which is between the said planes ; and the segment cut
off from the cylinder is one sixth part of the whole prism,
2. If in a cube a cylinder be inscribed which has its
bases in the opposite parallelograms -f• and touches with its
surface the remaining four planes (faces), and if there also
be inscribed in the same cube another cylinder which has
its bases in other parallelograms and touches with its surface
the remaining four planes (faces), then the figure bounded
by the surfaces of the cylinders, which is within both cylinders,
is two-thirds of the whole cube.
Now these theorems differ in character from those commu-
nicated before ; for we compared the figures then in question,
* The parallelograms are apparently sgwares. f i.e. squares.
THE METHOD 13
conoids and spheroids and segments of them, in respect of size,
with figures of cones and cylinders : but none of those figures
have yet been found to be equal to a solid figure bounded by
planes; whereas each of the present figures bounded by two
planes and surfaces of cylinders is found to be equal to one of
the solid figures which are bounded by planes. The proofs then
of these theorems I have written in this book and now send
to you. Seeing moreover in you, as I say, an earnest student,
a man of considerable eminence in philosophy, and an admirer
[of mathematical inquiry], I thought fit to write out for you
and explain in detail in the same book the peculiarity of a
certain method, by which it will be possible for you to get
a start to enable you to investigate some of the problems in
mathematics by means of mechanics. This procedure is, I am
persuaded, no less useful even for the proof of the theorems
themselves ; for certain things first became clear to me by a
mechanical method, although they had to be demonstrated by
geometry afterwards because their investigation by the said
method did not furnish an actual demonstration. But it is of
course easier, when we have previously acquired, by the method,
some knowledge of the questions, to supply the proof than
it is to find it without any previous knowledge. This is a
reason why, in the case of the theorems the proof of which
Eudoxus was the first to discover, namely that the cone is
a third part of the cylinder, and the pyramid of the prism,
having the same base and equal height, we should give no
small share of the credit to Democritus who was the first
to make the assertion with regard to the said figure* though
he did not prove it. I am myself in the position of having
first made the discovery of the theorem now to be published
[by the method indicated], and I deem it necessary to expound
the method partly because I have already spoken of itf and
I do not want to be thought to have uttered vain words, but
* irepl του βίρ-ημένον σχήματος, in the singular. Possibly Archimedes may
have thought of the case of the pyramid as being the more fundamental and as
really involving that of the cone. Or perhaps " figure " may be intended for
" type of figure."
t Of. Preface to Quadrature of Parabola.
14 ARCHIMEDES
equally because I am persuaded that it will be of no little
service to mathematics ; for I apprehend that some, either of
my contemporaries or of my successors, will, by means of the
method when once established, be able to discover other
theorems in addition, which have not yet occurred to me.
First then I will set out the very first theorem which
became known to me by means of mechanics, namely that
Any segment of a section of a right-angled cone {i.e. a parabola)
is four-thirds of the triangle which has the same base and equal
height,
and after this I will give each of the other theorems investi-
gated by the same method. Then, at the end of the book,
I will give the geometrical [proofs of the propositions]...
[I premise the following propositions which I shall use
in the course of the work.]
1. If from [one magnitude another magnitude be sub-
tracted which has not the same centre of gravity, the centre
of gravity of the remainder is found by] producing [the
straight line joining the centres of gravity of the whole
magnitude and of the subtracted part in the direction of
the centre of gravity of the whole] and cutting off from it
a length which has to the distance between the said centres
of gravity the ratio which the weight of the subtracted
magnitude has to the weight of the remainder.
\0n the Equilibrium of Planes, i. 8]
2. If the centres of gravity of any number of magnitudes
whatever be on the same straight line, the centre of gravity
of the magnitude made up of all of them will be on the same
straight line. [C£ Ibid. i. 5]
3. The centre of gravity of any straight line is the point
of bisection of the straight line. [Cf Ibid. i. 4]
4. The centre of gravity of any triangle is the point in
which the straight lines drawn from the angular points of
the triangle to the middle points of the (opposite) sides cut
one another. \_Ibid. i. 13, 14]
5. The centre of gravity of any parallelogram is the point
in which the diagonals meet. [^Ibid. i. 10]
THE METHOD 15
6. The centre of gravity of a circle is the point which is
also the centre [of the circle].
7. The centre of gravity of any cylinder is the point of
bisection of the axis.
8. The centre of gravity of any cone is [the point which
divides its axis so that] the portion [adjacent to the vertex is]
triple [of the portion adjacent to the base].
[All these propositions have already been] proved *. [Besides
these I require also the following proposition, which is easily
proved :
If in two series of magnitudes those of the first series are,
in order, proportional to those of the second series and further]
the magnitudes [of the first series], either all or some of them,
are in any ratio whatever [to those of a third series], and if the
magnitudes of the second series are in the same ratio to the
corresponding magnitudes [of a fourth series], then the sum
of the magnitudes of the first series has to the sum of the
selected magnitudes of the third series the same ratio which
the sum of the magnitudes of the second series has to the
sum of the (correspondingly) selected magnitudes of the fourth
series. [On Conoids and Spheroids, Prop. 1.] "
Proposition 1 .
Let ABC be a segment of a parabola bounded by the
straight line AC and the parabola ABC, and let D be the
middle point of AC. Draw the straight line DBE parallel
to the axis of the parabola and join AB, BC.
Then shall the segment ABC be | of the triangle ABC.
From A draw AKF parallel to BE, and let the tangent
to the parabola at G meet DBE in Ε and AKF in F. Produce
CB to meet AF in K, and again produce CK to H, making
KH equal to CK.
* The problem of finding the centre of gravity of a cone is not solved in
any extant work of Archimedes. It may have been solved either in a separate
treatise, such as the irepl ^vyCiv, which is lost, or perhaps in a larger mechanical
work of which the extant books On the Equilibrium of Planes formed only a part.
16
ARCHIMEDES
Consider GH as the bar of a balance, Κ being its middle
point.
Let MO be any straight line parallel to ED, and let it meet
OF, CK, ΛΟίϋΜ,Ν,Ο and the curve in P.
Now, since CE is a tangent to the parabola and CD the
semi-ordinate,
EB = BD;
" for this is proved in the Elements [of Conies] *."
Since FA, MO are parallel to EB, it follows that
FK = KA, MN=NO.
Now, by the property of the parabola, " proved in a lemma,"
MO :OP = GA:AO [Cf. Quadrature of Parabola, Prop. 5]
= GK : KN [Eucl. vi. 2]
= HK:KN.
Take a straight line TG equal to OP, and place it with its
centre of gravity at H, so that TH = HG ; then, since Ν is the
centre of gravity of the straight line MO,
and MO:TG = HK: KN,
* i.e. the works on conies by Aristaeus and Euclid. Cf. the similar
expression in On Conoids and Spheroids, Prop. 3, and Quadrature of Parabola,
Prop. 3.
λ
THE METHOD 17
it follows that TG at Η and MO at Ν will be in equilibrium
about K. [On the Equilihriwm of Planes, i. 6, 7]
Similarly, for all other straight lines parallel to BE and
meeting the arc of the parabola, (1) the portion intercepted
between FG, AG with its middle point on KG and (2) a
length equal to the intercept between the curve and AG
placed with its centre of gravity at Η will be in equilibrium
about K.
Therefore Κ is the centre of gravity of the whole system
consisting (1) of all the straight lines as MO intercepted between
FG, AG and placed as they actually are in the figure and (2) of
all the straight lines placed at Η equal to the straight lines
as PO intercepted between the curve and AG.
And, since the triangle GFA is made up of all the parallel
lines like MO,
and the segment GBA is made up of all the straight lines like
PO within the curve,
it follows that the triangle, placed where it is in the figure, is
in equilibrium about Κ with the segment GBA placed with its
centre of gravity at H.
Divide KG at W so that GK=^KW\
then W is the centre of gravity of the triangle A GF ; " for this
is proved in the books on equilibrium " {ev τοί? ΙσορροΐΓίκοΐ<ΐ).
[Cf. On the Equilibrium of Planes I. 15]
Therefore l^.AGF : (segment ABG) = HK : KW
= 3:1.
Therefore segment ABG=^AAGF.
But AAGF = 4>AABG.
Therefore segment ABG = ^AABG.
"Now the fact here stated is not actually demonstrated
by the argument used; but that argument has given a sort
of indication that the conclusion is true. Seeing then that
the theorem is not demonstrated, but at the same time
H.A. 2
18 ARCHIMEDES
suspecting that the conclusion is true, we shall have recourse
to the geometrical demonstration which I myself discovered
and have already published*."
Proposition 2.
We can investigate by the same method the propositions that
(1) Any sphere is {in respect of solid content) four times
the cone with base equal to a great circle of the sphere and
height equal to its radius; and
(2) the cylinder with base equal to a great circle of the
sphere and height equal to the diameter is 1^ ti7nes the sphere.
(1) Let ABCD be a great circle of a sphere, and AG, BD
diameters at right angles to one another.
Let a circle be drawn about BD as diameter and in a plane
perpendicular to AG, and on this circle as base let a cone
be described with A as vertex. Let the surface of this cone
be produced and then cut by a plane through G parallel to
its base ; the section will be a circle on EF as diameter. On
this circle as base let a cylinder be erected with height and
axis AG, and produce GA to H, making AH equal to GA.
Let GH be regarded as the bar of a balance, A being its
middle point.
Draw any straight line MN in the plane of the circle
ABGD and parallel to BD. Let MN meet the circle in 0, P,
the diameter AG in β, and the straight lines AE, AF in Q, R
respectively. Join AG.
* The word governing την 'γεωμετρονμάΐ'ην άττόδειξιν in the Greek text is
τάξομ€ΐ>, a reading which seems to be doubtful and is certainly difficult to
translate. Heiberg translates as if τάξομεν meant "we shall give lower down"
or "later on," but I agree with Th. Beinach {Revue generate des sciences pures
et appliquees, 30 November 1907, p. 918) that it is questionable whether
Archimedes would really have written out in full once more, as an appendix,
a proof which, as he says, had already been published (i.e. presumably in the
Quadrature of a Parabola), τάξομεν, if correct, should apparently mean "we
shall appoint," "prescribe" or "assign."
THE METHOD
19
Through MN draw a plane at right angles to AG;
this plane will cut the cylinder in a circle with diameter MN,
the sphere in a circle with diameter OP, and the cone in a
circle with diameter QR.
Now, since MS = AC, and Q8=AS,
MS.SQ = CA.AS
= A0'
Μ
L V
Η
A X G
^
\X
a
/O /Q
S R\P\
D
V
κ
w
And, since HA = AG,
HA:AS=CA:AS
= MS:SQ
= M8':MS.SQ
= MS' : {OS' + SQ% from above,
= MN' : (OP' + QR')
= (circle, diam. MN) : (circle, diam. OP
+ circle, diam. QR).
That is,
HA : AS = (circle in cylinder) : (circle in sphere + circle in cone).
Therefore the circle in the cylinder, placed where it is, is
in equilibrium, about A, with the circle in the sphere together
2—2
20 ARCHIMEDES
with the circle in the cone, if both the latter circles are placed
with their centres of gravity at H.
Similarly for the three corresponding sections made by a
plane perpendicular to AG and passing through any other
straight line in the parallelogram LF parallel to EF.
If we deal in the same way with all the sets of three circles
in which planes perpendicular to AG cut the cylinder, the
sphere and the cone, and which make up those solids respec-
tively, it follows that the cylinder, in the place where it is, will
be in equilibrium about A with the sphere and the cone together,
when both are placed with their centres of gravity at H.
Therefore, since Κ is the centre of gravity of the cylinder,
HA : -4^ = (cylinder) : (sphere + cone AEF).
BvitHA=2AK;
therefore cylinder = 2 (sphere + cone AEF).
Now cylinder = 3 (cone AEF) ; [Eucl. xii. 10]
therefore cone AEF = 2 (sphere).
But, since EF=2BD,
cone AEF =8 (cone ABB) ;
therefore sphere = 4 (cone ABD).
(2) Through B, D draw VBW, XDY parallel to ^C;
and imagine a cylinder which has AG for axis and the circles
on VX, TTF as diameters for bases.
Then cylinder FF= 2 (cylinder FD)
= 6 (cone ABD) [Eucl. xii. 10]
= f (sphere), from above.
Q.E.D.
"From this theorem, to the effect that a sphere is four
times as great as the cone with a great circle of the sphere as
base and with height equal to the radius of the sphere, I con-
ceived the notion that the surface of any sphere is four times as
great as a great circle in it; for, judging from the fact that any
circle is equal to a triangle with base equal to the circumference
and height equal to the radius of the circle, I apprehended
THE METHOD 21
that, in like maimer, any sphere is equal to a cone with base
equal to the surface of the sphere and height equal to the
radius*."
Proposition 3.
By this method we can also investigate the theorem that
A cylinder with base equal to the greatest circle in a spheroid
and height equal to the axis of the spheroid is 1\ times the
spheroid;
and, when this is established, it is plain that
If any spheroid he cut by a plane through the centre and at
right angles to the axis, the half of the spheroid is double of the
cone which has the same base and the same axis as the segment
{i.e. the half of the spheroid).
Let a plane through the axis of a spheroid cut its surface in
the ellipse A BCD, the diameters (i.e. axes) of which are AG,
BD ; and let Κ be the centre.
DraAv a circle about BD as diameter and in a plane per-
pendicular to AG]
imagine a cone with this circle as base and A as vertex
produced and cut by a plane through G parallel to its base ;
the section will be a circle in a plane at right angles to AG
and about EF as diameter.
Imagine a cylinder with the la^"ter circle as base and axis
AG\ produce GA to H, making AH equal to GA.
Let HG be regarded as the bar of a balance, A being its
middle point.
In the parallelogram LF draw any straight line MN
parallel to EF meeting the ellipse in 0, Ρ and AE, AF, AG m
Q, R, S respectively.
* That is to say, Archimedes originally solved the problem of finding the
solid content of a sphere before that of finding its surface, and he inferred the
result of the latter problem from that of the former. Yet in On the Sphere and
Cylinder i. the surface is independently found (Prop. 33) and before the
volume, which is found in Prop. 34 : another illustration of the fact that the
order of propositions in the treatises of the Greek geometers as finally
elaborated does not necessarily follow the order of discovery.
22
ARCHIMEDES
If now a plane be drawn through MN at right angles to
AC, it will cut the cylinder in a circle with diameter MN, the
spheroid in a circle with diameter OP, and the cone in a circle
with diameter QR.
^mQeHA = AG,
Therefore
HA
HA
A8=CA
= EA
= MS
AS
AQ
SQ.
AS = MS':MS.8Q.
Η
Ε W C Υ .
But, by the property of the ellipse,
A8.SG:S0' = AK':KB'
= A8' :8Q';
therefore 8Q^ : 80^ = A8' : A8 . 80
= 8Q' :8Q .QM,
and accordingly 80^ = 8Q . QM.
Add 8Q'^ to each side, and we have
80' + 8Q' = 8Q.8M.
Therefore, from above, we have
HA:A8=M8' :(80' +8Q')
= MN':(OP' + QR^
= (circle, diam. ilOT): (circle, diara. OP + circle, diam. QR)
THE METHOD 23
That is,
Ξ A :AS= (circle in cylinder) : (circle in spheroid + circle in cone).
Therefore the circle in the cylinder, in the place where it is,
is in equilibrium, about A, with the circle in the spheroid and
the circle in the cone together, if both the latter circles are
placed with their centres of gravity at H.
Similarly for the three corresponding sections made by a
plane perpendicular to AG and passing through any other
straight line in the parallelogram LF parallel to EF.
If we deal in the same way with all the sets of three circles
in which planes perpendicular to AG cut the cylinder, the
spheroid and the cone, and which make up those figures
respectively, it follows that the cylinder, in the place where it
is, will be in equilibrium about A with the spheroid and the
cone together, when both are placed with their centres of
gravity at H.
Therefore, since Κ is the centre of gravity of the cylinder,
HA : ^iT^ (cylinder) : (spheroid + cone AEF).
But HA = 2AK;
therefore cylinder = 2 (spheroid + cone AEF).
And cylinder = 3 (cone AEF) ; [Eucl. xii. 10]
therefore cone AEF = 2 (spheroid).
But, since ^i^=25i),
cone AEF = 8 (cone ABD) ;
therefore spheroid = 4 (cone ABD),
and half the spheroid = 2 (cone ABD).
Through B, D draw VBW, XDY parallel to AG;
and imagine a cylinder which has AG for axis and the circles
on VX, WY as diameters for bases.
Then cylinder FF=2 (cylinder VD)
= 6 (cone ABD)
= f (spheroid), from above.
Q.E.D.
24
ARCHIMEDES
Proposition 4.
Any segment of a right-angled conoid {i.e. a paraboloid of
revolution) cut off by a plane at right angles to the axis is
1\ times the cone which has the same base and the same axis
as the segment.
This can be investigated by our method, as follows.
Let a paraboloid of revolution be cut by a plane through
the axis in the parabola BAG]
and let it also be cut by another plane at right angles to the
axis and intersecting the former plane in BG. Produce DA,
the axis of the segment, to H, making HA equal to AD.
Μ
Ε
A F
Ο/ /
s\ \P
/ "
Κ \ \
Imagine that HD is the bar of a balance, A being its
middle point.
The base of the segment being the circle on BG as diameter
and in a plane perpendicular to AD,
imagine (1) a cone drawn with the latter circle as base and A
as vertex, and (2) a cylinder with the same circle as base and
AD as axis.
In the parallelogram EG let any straight line MN be drawn
parallel to BG, and through MN let a plane be drawn at right
angles to AD\ this plane will cut the cylinder 'in a circle with
diameter MN and the paraboloid in a circle with diameter OF.
THE METHOD 25
Now, Β A C being a parabola and BD, OS ordinates,
ΌΛ :AS = BD':OS',
or HA:AS = M8\:S0\
Therefore
HA : ^>Si = (circle, rad. MS) : (circle, rad. OS)
= (circle in cylinder) : (circle in paraboloid).
Therefore the circle in the cylinder, in the place where it is,
will be in equilibrium, about A, with the circle in the paraboloid,
if the latter is placed with its centre of gravity at H.
Similarly for the two corresponding circular sections made
by a plane perpendicular to AD and passing through any other
straight line in the parallelogram which is parallel to BC.
Therefore, as usual, if we take all the circles making up the
whole cylinder and the whole segment and treat them in the
same way, we find that the cylinder, in the place where it is,
is in equilibrium about A with the segment placed with its
centre of gravity at H.
If Κ is the middle point oi AD, Κ is the centre of gravity
of the cylinder ;
therefore HA : ^Z'= (cylinder) : (segment).
Therefore cylinder = 2 (segment).
And cylinder = 3 (cone ^5C); [Eucl. xii. 10]
therefore segment = f (cone ABC).
Proposition 5.
The centre of gravity of a segment of a right-angled conoid
{i.e. a paraboloid of revolution) cut off by a plane at right angles
to the axis is on the straight line which is the axis of the segment,
and divides the said straight line in such a way that the portion
of it adjacent to the vertex is double of the remaining portion.
This can be investigated by the method, as follows.
Let a paraboloid of revolution be cut by a plane through
the axis in the parabola BAG ;
and let it also be cut by another plane at right angles to the
axis and intersecting the former plane in BG.
26
ARCHIMEDES
Produce ΏΛ, the axis of the segment, to H, making HA
equal to AD; and imagine Ό Η to be the bar of a balance, its
middle point being A.
The base of the segment being the circle on BG as diameter
and in a plane perpendicular to AD,
imagine a cone with this circle as base and A as vertex, so that
AB, AG are generators of the cone.
In the parabola let any double ordinate OP be drawn
meeting AB, AD, AG in Q, S, R respectively.
Β
D C
Now, from th€
} property of the parabola.
BD':08' = DA : AS
= BD :QS
= BD':BD.Q8.
Therefore
OS'=BD.QS,
or
BD:OS=OS :QS,
whence
BD:QS=OS':QS\
But
BD:QS=AD:AS
:=HA:AS.
Therefore
HA :A8=0S':QS'
= OP':QR\
THE METHOD 27
If now through OP ά plane be drawn at right angles to
AD, this plane cuts the paraboloid in a circle with diameter
OP and the cone in a circle with diameter QR.
We see therefore that
HA : J.^= (circle, diam. OP) : (circle, diam. QR)
= (circle in paraboloid) : (circle in cone);
and the circle in the paraboloid, in the place where it is, is in
equilibrium about A with the circle in the cone placed with its
centre of gravity at S.
Similarly for the two corresponding circular sections made
by a plane perpendicular to AD and passing through any other
ordinate of the parabola.
Dealing therefore in the same way with all the circular
sections which make up the whole of the segment of the
paraboloid and the cone respectively, we see that the segment
of the paraboloid, in the place where it is, is in equilibrium
about A with the cone placed with its centre of gravity at H.
Now, since A is the centre of gravity of the whole system
as placed, and the centre of gravity of part of it, namely the
cone, as placed, is at H, the centre of gravity of the rest,
namely the segment, is at a point Κ on HA produced such
that
HA : AK = (segment) : (cone).
But segment = f (cone). [Prop. 4]
Therefore HA = ^AK;
that is, Κ divides AD in such a way that AK= 2KD.
Proposition 6.
The centre of gravity of any hemisphere [is on the Straight
line which] is its axis, and divides the said straight line in such
a way that the portion of it adjacent to the surface of the
hemisphere has to the remaining portion the ratio which 5 has
toS.
Let a sphere be cut by a plane through its centre in the
circle ABCD ;
28
ARCHIMEDES
let AG, BD be perpendicular diameters of this circle,
and through BD let a plane be drawn at right angles to AG.
The latter plane will cut the sphere in a circle on BD as
diameter.
Imagine a cone with the latter circle as base and A as
vertex.
Produce GA to H, making AH equal to GA, and let HG be
regarded as the bar of a balance, A being its middle point.
In the semicircle BAD, let any
straight line OP be drawn parallel to
BD and cutting AG in Ε and the two
generators AB, AD of the cone in Q, R
respectively. Join AG.
Through OP let a plane be drawn
at right angles to AG;
this plane will cut the hemisphere in a
circle with diameter OP and the cone
in a circle with diameter QR.
Now
HA'.AE = AG:AE
= AO':AE'
= {OE'+AE'):AE'
= {OE' + QEy.QE'
= (circle, diam. OP + circle, diam. QR) : (circle, diam. QR).
Therefore the circles with diameters OP, QR, in the places
where they are, are in equilibrium about A with the circle with
diameter QR if the latter is placed with its centre of gravity
at H.
And, since the centre of gravity of the two circles with
diameters OP, QR taken together, in the place where they are,
is
[There is a lacuna here ; but the proof can easily be com-
pleted on the lines of the corresponding but more difficult case
in Prop. 8.
We proceed thus from the point where the circles with
diameters OP, QR, in the place where they are, balance, about A,
THE METHOD 29
the circle with diameter QR placed with its centre of gravity
at H.
A similar relation holds for all the other sets of circular
sections made by other planes passing through points on AG
and at right angles to AG.
Taking then all the circles which fill up the hemisphere
BAD and the cone ABD respectively, we find that
the hemisphere BAD and the cone ABD, in the places where
they are, together balance, about A, a cone equal to ABD placed
with its centre of gravity at H.
Let the cylinder Μ + Ν he equal to the cone ABD.
Then, since the cylinder ilf+iV placed with its centre of
gravity at Η balances the hemisphere BAD and the cone ABD
in the places where they are,
suppose that the portion Μ of the cylinder, placed with its
centre of gravity at H, balances the cone ABD (alone) in the
place where it is; therefore the portion Ν of the cylinder placed
with its centre of gravity at Η balances the hemisphere (alone)
in the place where it is.
Now the centre of gravity of the cone is at a point V such
Ui^tAG = ^GV]
therefore, since ilf at if is in equilibrium with the cone,
Μ : (cone) = ^AG : HA^^AC : AC,
whence ^ = f (cone).
But M+N= (cone) ; therefore iV = | (cone).
Now let the centre of gravity of the hemisphere be at W,
which is somewhere on AG.
Then, since Ν at Η balances the hemisphere alone,
(hemisphere) : N = HA : AW.
But the hemisphere BAD = twice the cone ABD;
[On the Sphere and Cylinder i. 34 and Prop. 2 above]
and iV = I (cone), from above.
Therefore 2:^ = HA : AW
= 2AG:AW,
whence AW = ^AG, so that W divides AG in such a way that
AW: WG = 5:S.]
30
ARCHIMEDES
Proposition 7.
We can also investigate by the same method the theorem
that
[Any segment of a sphere has] to the cone [with the same
base and height the ratio which the sum of the radius of the
sphere and the height of the complementary segment has to
the height of the complementary segment]
[There is a lacuna here; but all that is missing is the
construction, and the construction is easily understood by
means of the figure. BAD is of course the segment of the
sphere the volume of which is to be compared with the volume
of a cone with the same base and height.]
The plane drawn through Μ Ν and at right angles to AG
will cut the cylinder in a circle with diameter MN, the segment
of the sphere in a circle with diameter OP, and the cone on
the base EF in a circle with diameter QR.
In the same way as before [cf Prop. 2] we can prove that
the circle with diameter MN, in the place where it is, is in
THE METHOD 31
equilibrium about Λ with the two circles with diameters OP,
QR if these circles are both moved and placed with their centres
of gravity at H.
The same thing can be proved of all sets of three circles
in which the cylinder, the segment of the sphere, and the
cone with the common height AG are all cut by any plane
perpendicular to AC.
Since then the sets of circles make up the whole cylinder,
the whole segment of the sphere and the whole cone respec-
tively, it follows that the cylinder, in the place where it is,
is in equilibrium about A with the sum of the segment of
the sphere and the cone if both are placed with their centres
of gravity at H.
Divide AG a-t W, F in such a way that
AW=WG, AV=SVG.
Therefore W will be the centre of gravity of the cylinder,
and V will be the centre of gravity of the cone.
Since, now, the bodies are in equilibrium as described,
(cylinder) : (cone AEF+ segment BAD of sphere)
= HA.AW.
[The rest of the proof is lost ; but it can easily be supplied
thus.
We have
(cone AEF+ segmt. BAD) : (cylinder) = AW : AG
= AW.AG'.AG\
But (cylinder) : (cone AEF) = AC': ^EG^
^AC':iAG'.
Therefore, ew aequali,
(cone AEF-^segmi.BAD) : (cone AEF) = AW . AG : ^AG^
= UG:iAG,
whence (segmt. BAD) : (cone AEF) = (^AG - ^ AG) : ^A G.
Again (cone AEF) : (cone ABD) = EG' : DG^
= AG':AG.GG
= AG:GG
= iAG:iGG.
32
ARCHIMEDES
Therefore, ex aequali,
(segment BAD) : (cone ABD) = {^AG -^AG): ^GG
= {^AG-AG):GG
= qAG+GC):GG.
Q.E.D.]
Proposition 8.
[The enunciation, the setting-out, and a few words of the
construction are missing.
The enunciation however can be supplied from that of
Prop. 9, with which it must be identical except that it cannot
refer to " ayiy segment," and the presumption therefore is that
the proposition was enunciated with reference to one kind of
segment only, i.e. either a segment greater than a hemisphere
or a segment less than a hemisphere.
Heiberg's figure corresponds to the case of a segment
greater than a hemisphere. The
segment investigated is of course
the segment BAD. The setting-
out and construction are self-
evident from the figure.]
Produce AC to H,G, making
HA equal to AG and GO equal
to the radius of the sphere ;
and let HG be regarded as the
bar of a balance, the middle point
being A.
In the plane cutting off the
segment describe a circle with G
as centre and radius {GE) equal
to AG; and on this circle as
base, and with A as vertex, let
a cone be described. AE, AF
are generators of this cone.
Draw KL, through any point
Q on AG, parallel to EF and cutting the segment in K, L, and
AE,AFmR,P respectively. Join AK.
THE METHOD 33
Now HA:AQ=GA:AQ
= AK^:AQ^
= {KQ^ + QA'^:qA^
= (circle, diam. KL + circle, diam. PR)
: (circle, diam. PR).
Imagine a circle equal to the circle with diameter PR
placed with its centre of gravity at H;
therefore the circles on diameters KL, PR, in the places where
they are, are in equilibrium about A with the circle with
diameter PR placed with its centre of gravity at H.
Similarly for the corresponding circular sections made by
any other plane perpendicular to AG.
Therefore, taking all the circular sections which make up
the segment ABD of the sphere and the cone AEF respec-
tively, we find that the segment ABD of the sphere and the
cone AEF, in the places where they are, are in equilibrium
with the cone A EF assumed to be placed with its centre of
gravity at H.
Let the cylinder If + JV be equal to the cone AEF which
has A for vertex and the circle on EF as diameter for base.
Divide ^G^ at Fso that ^(7 = 4F(^;
therefore V is the centre of gravity of the cone AEF; "for
this has been proved before*."
Let the cylinder Μ + Ν he cut by a plane perpendicular to
the axis in such a way that the cylinder Μ (alone), placed with
its centre of gravity at H, is in equilibrium with the cone AEF.
Since Μ -\- Ν suspended at Η is in equilibrium with the
segment ABD of the sphere and the cone AEF in the places
where they are,
while M, also at H, is in equilibrium with the cone AEF in
the place where it is, it follows that
iV at -ff is in equilibrium with the segment ABD of the
sphere in the place where it is.
* Cf. note on p. 15 above,
TT A. 3
34 ARCHIMEDES
Now (segment ABD of sphere) : (cone ABB)
^OG:GG;
"for this is already proved" [Cf. On the Sphere and Cylinder
II, 2 Cor. as well as Prop. 7 ante\
And (cone ABD) : (cone AEF)
= (circle, diam. BD) : (circle, diam. EF)
= ΒΌ' : EF'
= BG' ; GE^
= CG.GA: GA'
= GG : GA.
Therefore, ex aequali,
(segment ABD of sphere) : (cone AEF)
= OG : GA.
Take a point W on AG such that
AW:WG = {GA+ 4>GG) : {GA + 2GG).
We have then, inversely,
GW: WA = (2GG + GA) : (4>GC + GA),
and, componendo,
GA:AW= {6GG + 2GA) : {4,GG + GA).
But GO = l (QGG + 2GA), [for GO - GG = ^ (GG + GA)]
and GV=l(^GG+GA);
therefore GA : AW = OG : GV,
and, alternately and inversely,
OG:GA = GV: WA.
It follows, from above, that
(segment ABD of sphere) : (cone AEF) = GV : WA.
Now, since the cylinder Μ with its centre of gravity at Η
is in equilibrium about A with the cone AEF with its centre
of gravity at V,
(cone AEF) : (cylinder M) = HA:AV
= CA:AV;
and, since the cone AEF = the cylinder M+N, we have,
dividendo and invertendo,
(cylinder M) : (cylinder N) = AV:GV.
THE METHOD 35
Hence, componendo,
(cone AEF) : (cylinder N)= GA : GV*
= HA:GV.
But it was proved that
(segment ABD of sphere) : (cone AEF) = GV : WA;
therefore, ex aequali,
(segment ABD of sphere) : (cylinder N) = HA : A W.
And it was above proved that the cylinder iV at Η is
in equilibrium about A with the segment ABB, in the place
where it is;
therefore, since Η is the centre of gravity of the cylinder N,
W is the centre of gravity of the segment ABB of the sphere.
Proposition 9.
In the same way we can investigate the theorem that
The centre of gravity of any segment of a sphere is on the
straight line which is the axis of the segment, and divides this
straight line in such a way that the part of it adjacent to the
vertex of the segment has to the remaining part the ratio which
the sum of the axis of the segment and four times the axis of
the complementary segment has to the sum. of the axis of the
segment and double the axis of the complementary segment.
[As this theorem relates to " any segment " but states the
same result as that proved in the preceding proposition, it
follows that Prop. 8 must have related to one kind of segment,
either a segment greater than a semicircle (as in Heiberg's
figure of Prop. 8) or a segment less than a semicircle; and
the present proposition completed the proof for both kinds of
segments. It would only require a slight change in the figure,
in any case.]
Proposition lO.
By this method too we can investigate the theorem that
\^A segment of an obtuse-angled conoid (i.e. a hyperboloid of
revolution) has to the cone which has] the same base [as the
* Archimedes arrives at this result in a very roundabout way, seeing that it
could have been obtained at once convertendo. Cf. Euclid x. 14.
3—2
36 ARCHIMEDES
segment and equal height the same ratio as the sum of the axis
of the segment and three times'] the " annex to the axis " {i.e. half
the transverse axis of the hyperbolic section through the axis of
the hyperholoid or, in other words, the distance between the
vertex of the segment and the vertex of the enveloping cone) has to
the sum of the axis of the segment and double of the " annex " *
[this is the theorem proved in On Conoids and Spheroids,
Prop. 25], " and also many other theorems, which, as the method
has been made clear by means of the foregoing examples, I will
omit, in order that I may now proceed to compass the proofs
of the theorems mentioned above."
Proposition 1 1 .
If in a right prism with square bases a cylinder be inscribed
having its bases in opposite square faces and touching tvith its
surface the remaining four parallelogrammic faces, and if
through the centre of the circle which is the base of the cylinder
and one side of the opposite square face a plane be drawn, the
figure cut off by the plane so drawn is one sixth part of the
whole prism.
"This can be investigated by the method, and, when it
is set out, I will go back to the proof of it by geometrical
considerations."
[The investigation by the mechanical method is contained
in the two Propositions, 11, 12. Prop. 13 gives another solution
which, although it contains no mechanics, is still of the character
which Archimedes regards as inconclusive, since it assumes that
the solid is actually made up of parallel plane sections and that
an auxiliary parabola is actually made up of parallel straight
lines in it. Prop. 14 added the conclusive geometrical proof]
Let there be a right prism with a cylinder inscribed as
stated.
* The text has "triple" (τριπλασίαν) in the last line instead of "double."
As there is a considerable lacuna before the last few lines, a theorem about the
centre of gravity of a segment of a hyperboloid of revolution may have fallen
out.
THE METHOD
37
u
G /
/
/
I
Let the prism be cut through the axis of the prism and
cylinder by a plane perpendicular
to the plane which cuts off the
portion of the cylinder; let this
plane make, as section, the paral-
lelogram ΛΒ, and let it cut the
plane cutting off the portion of
the cylinder (which plane is per-
pendicular to ΛΒ) in the straight
line BG.
Let CD be the axis of the
prism and cylinder, let EF bisect
it at rierht angles, and throuefh
EF let a plane be drawn at right
angles to CD; this plane will cut the prism in a square and
the cylinder in a circle.
Let MN be the square and OPQR the circle, and let the
circle touch the sides of the square
in 0, P, Q, R [F, Ε in the first
figure are identical with 0, Q
respectively]. Let £[ be the centre
of the circle.
Let KL be the intersection of
the plane through EF perpen-
dicular to the axis of the cylinder
and the plane cutting ofi" the
portion of the cylinder; KL is
bisected by OHQ [and passes
through the middle point of HQ].
Let any chord of the circle, as ST, be drawn perpendicular
to HQ, meeting HQ in W;
and through ST let a plane be drawn at right angles to OQ
and produced on both sides of the plane of the circle OPQR.
The plane so drawn will cut the half cylinder having the
semicircle PQR for section and the axis of the prism for height
in a parallelogram, one side of which is equal to ST and another
is a generator of the cylinder; and it will also cut the portion
y^'^^ L
\
r
Η
V
/
W
3
Μ
38 ARCHIMEDES
of the cylinder cut off in a parallelogram, one side of which is
equal to ST and the other is equal and parallel to ?7F (in the
first figure).
UV will be parallel to 5 F and will cut off, along UG in the
parallelogram DE, the segment EI equal to QW.
Now, since EG is a parallelogram, and VI is parallel to GG,
EG:GI=YG:GV
= BY:UV
= (ZZ7 in half cyl.) : (ZZ7 in portion of cyl.).
And EG = HQ, GI = HW, QH=OH;
therefore OH : HW={CJ in half cyl.) : (ZZ7 in portion).
Imagine that the parallelogram in the portion of the
cylinder is moved and placed at so that is its centre
of gravity, and that OQ is the bar of a balance, Η being its
middle point.
Then, since W is the centre of gravity of the parallelogram
in the half cylinder, it follows from the above that the paral-
lelogram in the half cylinder, in the place where it is, with its
centre of gravity at W, is in equilibrium about Η with the
parallelogram in the portion of the cylinder when placed with
its centre of gravity at 0.
Similarly for the other parallelogrammic sections made by
any plane perpendicular to OQ and passing through any other
chord in the semicircle PQR perpendicular to OQ.
If then we take all the parallelograms making up the half
cylinder and the portion of the cylinder respectively, it follows
that the half cylinder, in the place where it is, is in equilibrium
about Η with the portion of the cylinder cut off when the
latter is placed with its centre of gravity at 0.
Proposition 12.
Let the parallelogram (square) MN perpendicular to the
axis, with the circle OPQR and its diameters OQ, PR, be
drawn separately.
THE METHOD
89
Join HO, HM, and through them draw planes at right
angles to the plane of the circle,
producing them on both sides of
that plane.
This produces a prism with
triangular section GEM and height
equal to the axis of the cylinder;
this prism is \ of the original
prism circumscribing the cylinder.
Let LK, UT be drawn parallel
to OQ and equidistant from it,
cutting the circle in K, T, RP
in *S', F, and GH, Η Μ in W, V respectively.
Through LK, UT draw planes at right angles to PR,
producing them on both sides of the plane of the circle ;
these planes produce as sections in the half cylinder PQR
and in the prism GHM four parallelograms in Λvhich the
heights are equal to the axis of the cylinder, and the other
sides are equal to KS, TF, LW, UV respectively
\^
\K
/w\
Η
\ "^κ
J
/V_
F /T
Μ
[The rest of the proof is missing, but, as Zeuthen says*,
the result obtained and the method of arriving at it are plainly
indicated by the above.
Archimedes Λvishes to prove that the half cylinder PQR, in
the place where it is, balances the prism GHM, in the place
where it is, about Η as fixed point.
He has first to prove that the elements (1) the parallelo-
gram with side = KS and (2) the parallelogram with side = LW,
in the places where they are, balance about S, or, in other
words that the straight lines SK, LW, in the places where
they are, balance about S.
Now (radius of circle OPQRy = SK' + SH\
or SL' = SK' + SW\
Therefore LS' -SW' = SK%
and accordingly {LS^-SW).LW = SK\
whence I {LS + SW) :\SK=SK: LW.
* Zeuthen in Bibliotheca Mathematica VII3, 1906-7, pp. 352-3.
40 ARCHIMEDES
And ^(LS + SW) is the distance of the centre of gravity
of LW from S,
while ^SK is the distance of the centre of gravity of SK from S.
Therefore SK and LW, in the places where they are,
balance about S.
Similarly for the corresponding parallelograms.
Taking all the parallelogrammic elements in the half
cylinder and prism respectively, we find that
the half cylinder PQR and the prism GHM, in the places
where they are respectively, balance about H.
From this result and that of Prop. 11 we can at once deduce
the volume of the portion cut off from the cylinder. For in
Prop. 11 the portion of the cylinder, placed with its centre of
gravity at 0, is shown to balance (about H) the half-cylinder
in the place where it is. By Prop. 12 we may substitute for
the half-cylinder in the place where it is the prism GHM
of that proposition turned the opposite way relatively to RP.
The centre of gravity of the prism as thus placed is at a point
(say Z) on HQ such that HZ= ^HQ.
Therefore, assuming the prism to be applied at its centre of
gravity, we have
(portion of cylinder) : (prism) = ^HQ : OH
= 2:3;
therefore (portion of cylinder) = | (prism GHM)
= I" (original prism).
Note. This proposition of course solves the problem of
finding the centre of gravity of a half cylinder or, in other
words, of a semicircle.
For the triangle GHM in the place where it is balances,
about H, the semicircle PQR in the place where it is.
If then X is the point on HQ which is the centre of
gravity of the semicircle,
IHO .{AGHM) = HX . (semicircle PQR),
or ^HO . HO' = HX.^7r. HO' ;
that is, HX = ^. HQ.]
07Γ
THE METHOD
41
/ Μ
"^^o
L
-^
Κ S ,
A Ε
D
Proposition 13.
Let there be a right prism with square bases, one of which
isABCD;
in the prism let a cylinder be
inscribed, the base of which is
the circle EFQH touching the
sides of the square ABCD in
E, F, G, H.
Through the centre and
through the side corresponding
to CO in the square face opposite
to ABCD let a plane be drawn;
this will cut off a prism equal
to J of the original prism and
formed by three parallelograms and two triangles, the triangles
forming opposite faces.
In the semicircle EFG describe the parabola which has
FK for axis and passes through E, G;
draw MN parallel to KF meeting GE in M, the parabola
in L, the semicircle in and CD in N.
Then MN .NL = NF^;
"for this is clear." [Cf Apollonius, Conies i. 11]
[The parameter is of course equal to GK or KF.^
Therefore MN : NL = GK' : LS'.
Through MN draw a plane at right angles to EG ;
this will produce as sections (1) in the prism cut off from the
whole prism a right-angled triangle, the base of which is MN,
while the perpendicular is perpendicular at Ν to the plane
ABCD and equal to the axis of the cylinder, and the hypo-
tenuse is in the plane cutting the cylinder, and (2) in the
portion of the cylinder cut off a right-angled triangle the base
of which is MO, while the perpendicular is the generator of
the cylinder perpendicular at to the. plane KN, and the
hypotenuse is
3—5
42 ARCHIMEDES
[There is a lacuna here, to be supplied as follows.
Since MN : NL = QK' : LS'
it follows that MN : ML = MN' : (MN' - LS')
= MN' : (MN' - MK')
= MN' : MO'.
But the triangle (1) in the prism is to the triangle (2) in
the portion of the cylinder in the ratio of MN' -.MO'.
Therefore
(Δ in prism) : (Δ in portion of cylinder)
= MN:ML
= (straight line in rect. DG) : (straight line in parabola).
We now take all the corresponding elements in the prism,
the portion of the cylinder, the rectangle DG and the parabola
BFG respectively ;]
and it will follow that
(all the As in prism) : (all the As in portion of cylinder)
= (all the str. lines in ZZ7 OG)
: (all the straight lines between parabola and EG).
But the prism is made up of the triangles in the prism,
[the portion of the cylinder is made up of the triangles in it],
the parallelogram DG of the straight lines in it parallel to KF,
and the parabolic segment of the straight lines parallel to KF
intercepted between its circumference and EG ;
therefore (prism) : (portion of cylinder)
= (£7 GD) : (parabolic segment EFG).
But CJ GD = ^ (parabolic segment EFG) ;
" for this is proved in my earlier treatise."
[Quadrature of Parabola]
Therefore prism = f (portion of cylinder).
If then we denote the portion of the cylinder by 2, the
prism is 3, and the original prism circumscribing the cylinder
is 12 (being 4 times the other prism) ;
therefore the portion of the cylinder = ^ (original prism).
Q.E.D.
THE METHOD 43
[The above proposition and the next are peculiarly interest-
ing for the fact that the parabola is an auxiliary curve introduced
for the sole purpose of analytically reducing the required
cubature to the known quadrature of the parabola.]
Proposition 14.
Let there be a right prism with square bases [and a
cylinder inscribed therein having its base in the square ABGD
and touching its sides at E, F, G, H;
let the cylinder be cut by a plane through EG and the side
corresponding to CD in the square face opposite to ABGD].
This plane cuts off from the prism a prism, and from the
cylinder a portion of it.
It can be proved that the portion of the cylinder cut off by
the plane is ^ of the whole prism.
But we will first prove that it is possible to inscribe in
the portion cut off from the cylinder, and to circumscribe about
it, solid figures made up of prisms which have equal height
and similar triangular bases, in such a way that the circum-
scribed figure exceeds the inscribed by less than any assigned
magnitude
But it was proved that
(prism cut off by oblique plane)
< f (figure inscribed in portion of cylinder).
ΝοΛν
(prism cut off) : (inscribed figure)
= Ο DG : (Os inscribed in parabolic segment);
therefore ZZ7 BG < f (Os in parabolic segment) :
which is impossible, since " it has been proved elsewhere " that
the parallelogram DG^ is | of the parabolic segment.
Consequently
not greater.
44 ARCHIMEDES
And (all the prisms in prism cut off)
: (all prisms in circurascr. figure)
= (all OS in CJ DG)
: (all Os in fig. circumscr. about parabolic segmt.) ;
therefore
(prism cut off) : (figure circumscr. about portion of cylinder)
= (O DG) : (figure circumscr. about parabolic segment).
But the prism cut off by the oblique plane is > f of
the solid figure circumscribed about the portion of the
cylinder
[There are large gaps in the exposition of this geometrical
proof, but the way in which the method of exhaustion was
applied/and the parallelism between this and other applications
of it, are clear. The first fragment shows that solid figures
made up of prisms were circumscribed and inscribed to the
portion of the cylinder. The parallel triangular faces of these
prisms were perpendicular to GE in the figure of Prop. 13 ;
they divided GE into equal portions of the requisite smallness ;
each section of the portion of the cylinder by such a plane was
a triangular face common to an inscribed and a circumscribed
right prism. The planes also produced prisms in the prism cut
off by the same oblique plane as cuts off the portion of the
cylinder and standing on GD as base.
The number of parts into which the parallel planes divided
GE was made great enough to secure that the circumscribed
figure exceeded the inscribed figure by less than a small
assigned magnitude.
The second part of the proof began with the assumption
that the portion of the cylinder is > -| of the prism cut off;
and this was proved to be impossible, by means of the use of
the auxiliary parabola and the proportion
MN : ML = MN' : MO'
which are employed in Prop. 13.
THE METHOD
45
We may supply the missing proof as follows*.
In the accompanying figure are represented (1) the first
Μ
element-prism circumscribed to the portion of the cylinder,
(2) two element-prisms adjacent to the ordinate OM, of which
that on the left is circumscribed and
that on the right (equal to the other)
inscribed, (3) the corresponding element-
prisms forming part of the prism cut
off(CC'GEDD') which is I of the original
prism.
In the second figure are shown
element-rectangles circumscribed and
inscribed to the auxiliary parabola,
which rectangles correspond exactly to
the circumscribed and inscribed element-
prisms represented in the first figure
(the length of GM is the same in both
figures, and the breadths of the element-
rectangles are the same as the heights of the element-prisms) ;
* It is right to mention that this has already been done by Th. Eeinach in
his version of the treatise ("Un Traito de Goometrie inedit d'Archimede " in
Revue generale des sciences pures et appliquees, 30 Nov. and 15 Dee. 1907) ;
but I prefer my own statement of the proof.
46 ARCHIMEDES
the corresponding element-rectangles forming part of the
rectangle GD are similarly shown.
For convenience we suppose that OE is divided into an
even number of equal parts, so that GK contains an integral
number of these parts.
For the sake of brevity we will call each of the two element-
prisms of which OM is an edge "el. prism (0)" and each of
the element-prisms of which MNN'_ is a common face " el.
prism {N)." Similarly we will use the corresponding abbrevia-
tions " el. rect. {L) " and " el. rect. {N) " for the corresponding
elements in relation to the auxiliary parabola as shown in
the second figure.
Now it is easy to see that the figure made up of all the
inscribed prisms is less than the figure made up of the circum-
scribed prisms by twice the final circumscribed prism adjacent
to FK, i.e. by twice " el. prism {N) " ; and, as the height of this
prism may be made as small as we please by dividing GK into
sufficiently small parts, it follows that inscribed and circum-
scribed solid figures made up of element-prisms can be drawn
differing by less than any assigned solid figure.
(1) Suppose, if possible, that
(portion of cylinder) > | (prism cut off),
or (prism cut off) < | (portion of cylinder).
Let (prism cut off) = f (portion of cylinder — X), say.
Construct circumscribed and inscribed figures made up of
element-prisms, such that
(circumscr. fig.) — (inscr. fig.) < X.
Therefore (inscr. fig.) > (circumscr. fig. —X),
and a fortiori > (portion of cyl. — X).
It follows that
(prism cut off) < | (inscribed figure).
Considering now the element-prisms in the prism cut off
and those in the inscribed figure respectively, we have
el. prism {N) : el. prism (0) = MN'' : .¥ 0^
= MN : ML [as in Prop. 13]
= el. rect. (N) : el. rect. (L).
THE METHOD 47
It follows that
S {el. prism (JV)| : S {el. prism (0)}
= 1 {el. rect. {N)\ : X {el. rect. (L)].
(There are really two more prisms and rectangles in the
first and third than there are in the second and fourth terms
respectively; but this makes no difference because the first
and third terms may be multiplied by a common factor as
n/{n — 2) without affecting the truth of the proportion. Cf.
the proposition from On Conoids and Spheroids quoted on p. 15
above.)
Therefore
(prism cut off) : (figure inscr. in portion of cyl.)
= (rect. GD) : (fig. inscr. in parabola).
But it was proved above that
(prism cut off) < f (fig. inscr. in portion of cyl.) ;
therefore (rect. GD) < f (fig. inscr. in parabola),
and, a fortiori,
(rect. GD) < f (parabolic segmt.) :
which is impossible, since
(rect. (rD) = I (parabolic segmt.).
Therefore
(portion of cyl.) is not greater than | (prism cut off).
(2) In the second lacuna must have come the beginning of
the next reductio ad absurdum demolishing the other possible
assumption that the portion of the cylinder is < f of the prism
cut off.
In this case our assumption is that
(prism cut off) > f (portion of cylinder) ;
and we circumscribe and inscribe figures made up of element-
prisms, such that
(prism cut off) > f (fig. circumscr. about portion of cyl.).
48 ARCHIMEDES
We now consider the element-prisms in the prism cut off
and in the circumscribed figure respectively, and the same
argument as above gives
(prism cut off) : (fig. circumscr. about portion of cyl.)
= (rect. GD) : (fig. circumscr. about parabola),
whence it follows that
(rect. GD) > I (fig. circumscribed about parabola),
and, a fortiori,
(rect. GD) > f (parabolic segment) :
which is impossible, since
(rect. (rZ)) = f (parabolic segmt.).
Therefore
(portion of cyl.) is not less than f (prism cut off).
But it was also proved that neither is it greater ;
therefore (portion of cyl.) = | (prism cut off)
= ^ (original prism).]
[Proposition 15.]
[This proposition, which is lost, would be the mechanical
investigation of the second of the two special problems
mentioned in the preface to the treatise, namely that of the
cubature of the figure included between two cylinders, each
of which is inscribed in one and the same cube so that its
opposite bases are in two opposite faces of the cube and its
surface touches the other four faces.
Zeuthen has shown how the mechanical method can be
applied to this case*.
In the accompanying figure VWYX is a section of the
cube by a plane (that of the paper) passing through the axis
BD of one of the cylinders inscribed in the cube and parallel
to two opposite faces.
The same plane gives the circle A BCD as the section of
the other inscribed cylinder with axis perpendicular to the
* Zeuthen in Bibliotheca Mathematica VII3, 1906-7, pp. 356-7.
THE METHOD
49
plane of the paper and extending on each side of the plane
to a distance equal to the radius of the circle or half the side
of the cube.
AG is the diameter of the circle which is perpendicular
to BD.
Join ΛΒ, AD and produce them to meet the tangent at G
to the circle in E, F.
Then EG = GF = GA.
Let LG be the tangent at A, and complete the rectangle
EFGL.
Η
Μ
^
^^
Β
^
/Q
S R\P\
D
V
Κ 1
w
Draw straight lines from A to the four corners of the
section in which the plane through BD perpendicular to A Κ
cuts the cube. These straight lines, if produced, Avill meet the
plane of the face of the cube opposite to A in four points
forming the four corners of a square in that plane with sides
equal to EF or double of the side of the cube, and we thus
have a pyramid with A for vertex and the latter square for
base.
Complete the prism (parallelepiped) with the same base
and height as the pyramid.
Draw in the parallelogram LF any straight line MN parallel
to EF, and through MN draw a plane at right angles to AG.
50 ARCHIMEDES
This plane cuts —
(1) the solid included by the two cylinders in a square with
side equal to OP,
(2) the prism in a square with side equal to MN, and
(3) the pyramid in a square with side equal to QR.
Produce GA to H, making HA equal to AC, and imagine
HG to be the bar of a balance.
Now, as in Prop. 2, since MS = AC, QS = AS,
MS.8Q=GA.AS
= A0'
= US'" + SQ\
Also
HA :AS=GA : AS
=-MS:SQ
= MS':MS.SQ
= MS•" : {OS^ + SQ% from above,
= MN^ : {OP' + QR'')
= (square, side MN) : (sq., side OP + sq., side QR).
Therefore the square with side equal to MN, in the place
where it is, is in equilibrium about A with the squares with
sides equal to OP, QR respectively placed with their centres of
gravity at H.
Proceeding in the same way with the square sections
produced by other planes perpendicular to AG, we finally
prove that the prism, in the place where it is, is in equilibrium
about A with the solid included by the two cylinders and the
pyramid, both placed with their centres of gravity at H.
Now the centre of gravity of the prism is at K.
Therefore HA : AK = (prism) : (solid + pyramid)
or 2:1= (prism) : (solid + ^ prism).
Therefore 2 (solid) + | (prism) = (prism).
It follows that
(solid included by cylinders) = | (prism)
= I (cube). Q.E.D.
THE METHOD 51
There is no doubt that Archimedes proceeded to, and
completed, the rigorous geometrical proof by the method of
exhaustion.
As observed by Prof. C. Juel (Zeuthen I.e.), the solid in the
present proposition is made up of 8 pieces of cylinders of the
tjrpe of that treated in the preceding proposition. As however
the two propositions are separately stated, there is no doubt
that Archimedes' proofs of them were distinct.
In this case A G would be divided into a very large number
of equal parts and planes would be drawn through the points
of division perpendicular to AC. These planes cut the solid,
and also the cube VY, in square sections. Thus we can inscribe
and circumscribe to the solid the requisite solid figures made
up of element-prisms and differing by less than any assigned
solid magnitude; the prisms have square bases and their
heights are the small segments of AC. The element-prism
in the inscribed and circumscribed figures which has the square
equal to OP"^ for base corresponds to an element-prism in the
cube which has for base a square with side equal to that of
the cube ; and as the ratio of the element-prisms is the ratio
OS^ : BK^, we can use the same auxiliary parabola, and work
out the proof in exactly the same way, as in Prop. 14.]
CAMBEIDGE : FEINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PEESS
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